Consider this initial-rate data at a certain temperature for the reaction descri
ID: 914948 • Letter: C
Question
Consider this initial-rate data at a certain temperature for the reaction described by : OH-(aq)
OCl-(aq) + I- (aq) --------------> OI-(aq) + Cl-(aq)
Trial [OCl-]0 (M) [I-]0 (M) [OH-]0 (M) Initial rated (M/s)
1 0.00161 0.00161 0.530 0.000335
2 0.00161 0.00301 0.530 0.000626
3 0.00279 0.00161 0.710 0.000433
4 0.00161 0.00301 0.880 0.000377
Determine the tate law and the value of the rate constant for this reaction.
Explanation / Answer
Answer –Given, reaction – OH-(aq) + OCl-(aq) + I- (aq) -----> OI-(aq) + Cl-(aq)
Assume rate law is - Rate = k [OH-]x [OCl-]y [I-]z
In this rate law there are x,y and z are the order with respect to OH-, OCl- and I-
Rate1 = k [OH-]1x [OCl-]1y [I-]1z
Rate2 = k [OH-]2x [OCl-]2y [I-]2z
Rate3 = k [OH-]3x [OCl-]3y [I-]3z
Rate4 = k [OH-]4x [OCl-]4y [I-]4z
Order with respect to I-
Rate2/ Rate1 = k [OH-]2x [OCl-]2y [I-]2z / k [OH-]1x [OCl-]1y [I-]1z
0.000626 / 0.000335 = (0.530)x /(0.530)x * (0.00161)y /(0.00161)y *(0.00301)z /(0.00161)z
1.86 = (1.86)z
So, z = 1
Order with respect to OH-
Rate4/ Rat2 = k [OH-]4x [OCl-]4y [I-]4z / k [OH-]2x [OCl-]2y [I-]2z
0.000377 / 0.000626 = (0.880)x /(0.530)x * (0.00161)y /(0.00161)y *(0.00301)z /(0.00301)z
0.602 = (1.66)x
So, x = -1
Order with respect to OCl-
Rate2/ Rate1 = k [OH-]2x [OCl-]2y [I-]2z / k [OH-]1x [OCl-]1y [I-]1z
0.000626 / 0.000335 = (0.530)-1 /(0.530)-1 * (0.00161)y /(0.00161)y *(0.00301)1 /(0.00161)1
1.86 = (1)y (1.86)
(1)y = 1.86 /1.86
=1
y = 0
So the order with respect to OH-, OCl- and I-are -1, 0 and 1 respectively.
Overall order of reaction = -1+ 0 +1 = 0
So, rate law
Rate = k [OH-]-1 [I-]
Now we need to put the values and calculate k
0.000335 Ms-1 = k (0.530 M)-1 *(0.00161)
0.000335 Ms-1 = k*0.00304
k = 0.000335 Ms-1 /0.00304
k = 0.110 M s-1